CHEM18.1 Experiment 7_Heterogenous Equilibria

Experiment 7: Heterogeneous Equilibria

Chua, Krystle Ashley

Flores, Ralph Marco

THEORETICAL FRAMEWORK

Theoretical Framework• A heterogeneous equilibrium results from a

reversible reaction involving reactants and products that are in different phases.

• A precipitation reaction is a reaction in which soluble ions in separate solutions are mixed together to form an insoluble compound that settles out of solution as a solid (called precipitate)

Theoretical Framework• Centrifugation is a process that involves the

use of the centrifugal force for the separation of mixtures with a centrifuge. It increases the effective gravitational force on a test tube so as to more rapidly and completely cause the precipitate to gather on the bottom of the tube.

Theoretical Framework• Centrifugation is a process that involves the

use of the centrifugal force for the separation of mixtures with a centrifuge. It increases the effective gravitational force on a test tube so as to more rapidly and completely cause the precipitate to gather on the bottom of the tube.

Theoretical Framework• The processes of dissolution and

precipitation are the reverse of each other. Taken together they form a dynamic equilibrium. Whenever a supersaturated solution forms, the equilibrium state will sooner or later be achieved by precipitation of a solid salt. Whenever an unsaturated solution is present in contact with a solid salt, the equilibrium state will sooner or later be achieved by dissolution of all or part of the solid salt.

Theoretical Framework• The comparison of solubility product constant,

Ksp, to the reaction quotient, Qsp, can be used to determine whether a precipitate will form with a given concentration of ions. Comparison Shift of Reaction Precipitation?

Qsp < Ksp Forward No Precipitation

Qsp = Ksp No shift; In equilibrium

No Precipitation

Qsp > Ksp Backward Precipitation

Theoretical Framework• Factors that affect the dissolution of precipitates

1. Temperature ( Temperature Solubility)

2. Addition of complexing agents. ( Complexing Agents Solubility)

Theoretical Framework



• Fractional precipitation is the process by which two aqueous substances in a solution are separated through the addition of a common ion, taking advantage of their different concentration needs in order to form a precipitate.

EXPERIMENTAL DATA

Part A: Precipitation

1. Label 4 test tubes A to D

2. Mix the following to each test tube:• A: 5 drops 0.1 M CaCl2 + 5 drops 1 M NH4OH + 15

drops distilled H2O• B: 5 drops 0.1 M CaCl2 + 5 drops 1 M NaOH + 15

drops distilled H2O• C: 5 drops 0.1 M CaCl2 + 5 drops 3 M NaOAc + 15

drops distilled H2O• D: 5 drops 0.1 M CaCl2 + 5 drops 3 M NH4OH + 15

drops 3 M NH4Cl

3. Centrifuge each test tube and observe the result.

4. Calculate the Ion Product of Ca(OH)2

Experimental Observations

Test Tube Mixture Observation

A 0.1 M CaCl2 + 1 M NH4OH No precipitation

B 0.1 M CaCl2 + 1 M NaOH White precipitate

C 0.1 M CaCl2 + 3 M NaOAc White precipitate

D 0.1 M CaCl2 + 3 M NH4OH + 3 M NH4Cl No precipitation

Theoretical Calculations

Given that the Ksp for Ca(OH)2 is 6.5 x 10-6. We calculate for the Ion Product Constant and prove the existence of precipitate.

Ca2+ + OH- → Ca(OH)2

M1V1 = M2V2

(0.1M)(5 drops) = M2(25 drops)

M2 = 0.02 M Ca2+

Test Tube A

NH4OH → NH4+ + OH-

M1V1=M2V2

(1 M NH4OH)(5 drops) = M2(25 drops)

M2 = 0.2 M NH4OH

NH4+ + OH- ↔ NH3 + H2O

To get OH-, which is represented as X,

Kb for NH3= [NH4+][OH-] / [NH3]

1.76 x 10-5 = (x)(x) / 0.2 M[OH-] = 1.87 x 10-3 M

Ion Product Constant = [Ca2+]

[OH-]

Ca2+ + 2 OH- → Ca(OH-)2

(0.02 M) (1.87 x 10-3 M)2 =

6.97 x 10-8

If Ksp > Qsp, then no precipitation will occur. Correct observation.

Test Tube BNa + + OH- → NaOH

M1V1 = M2V2

(1 M NaOH)(5 drops) = M2(25 drops)M2 = 0.2 M NaOH

Ion Product Constant = [Ca2+][OH-]

Ca2+ + 2OH- → Ca(OH-)2

(0.02 M)(0.2 M)2 = 8 x 10-4

If Ksp < Qsp, then precipitation will occur.

Correct observation.

Test Tube C

NaOAc → Na + + OAc -

M1V1 = M2V2

(3 M NaOAc)(5 drops)=M2(25 drops)M2 = 0.6 M NaOAc

OAc - + H2O ↔ HOAc + OH-

Since Kw = KaKb ; where Kb = 1.0 x 10-14 / 1.8 x 10-5

5.56 x 10-10 = (x)(x) / 0.06[OH -] = 1.83 x 10-5

Ion Product Constant = [Ca2+][OH-]

Ca2+ + 2OH- → Ca(OH-)2

(0.02 M)(1.83 x 10-5)2 = 6.69 x 10-12

If Ksp > Qsp, then precipitation will occur. Correct observation.

Test Tube D

NH4OH → NH4+ + OH-

NH3 + H2O ↔ NH4+ + OH-

M1V1 = M2V2

(3 M)(5 drops) = M2(25 drops)M2 = 0.6 M NH4OH

NH4Cl → NH4

+ + Cl-M1V1 = M2V2

(4 M)(15 drops) = M2(25 drops)M2 = 2.4 M NH4

+

Kb of NH3 = (x)(2.4+x) / (0.6-x)

1.8 x 10-5 = 2.4x + x 2 / 0.6-x[OH-] = 4.49 x 10-6

Ion Product Constant = [Ca2+ ][OH-]

Ca2+ + 2OH- → Ca(OH-)2

(0.02 M)(4.49 x 10 -13) 2 = 4.05 x 10-13

If Ksp > Qsp, then no precipitation will occur. Correct observation.

PART B: Dissolution of Precipitates

Put 10 drops of:

• Pb(NO3) to test tube A• Ag(NO3) to test tube B

Add HCl dropwise to both test tubes until no more ppt

Centrifuge

Decant supernatant fluid and wash with 5 drops H2O

Add 2mL H2O shake!

Heat in warm bath, observe

Cool to room temperature, observe

Repeat every procedure from the start

• PbNO3 to test tube A• Ag(NO3)2 to test tube B

Add 2mL of conc. NH4OH gradually to the new test tubes, observe

Place 10 drops of

Add 3 drops of 0.1M K2CrO4 and 1M HNO3

Neutralize (add 3 drops 6M NaOH)

Centrifuge

Decant liquid

Add 6M NaOH, observe

Put 10 drops of:

• Pb(NO3) + 0.1 M Disodium carbonate to test tube A• Ag(NO3) + 0.1 M Disodium sulfate to test tube B

Observe

Discard liquid + 5 drops HCl

Put 10 drops Cu(NO3)2 to two test tubes

+ 3M HOAc until acidic

+ thioacetamide sol’n until full precipitation

Centrifuge

Discard liquid and + 2mL H2O

Add HCl in test tube A and HNO3 in test tube B

Heat in water bath

Observe

Experimental Results

• There was formation of precipitates for lead nitrate (white-colored) and silver nitrate (purple-colored)

• Reactions involved:

Pb2+ (aq) + 2Cl- (aq) PbCl2 (s)

Ag+ (aq)+ Cl- (aq) AgCl (s)• After heating, some of the lead chloride dissolved in the

solution while silver chloride remained the same


• After cooling back to room temperature, lead chloride came back to original amount

• After adding NH4OH to the precipitate, other precipitates formed: Pb(OH)2, [Ag(NH3)2]+reactions involved (w/ ammonia):

• Pb2+ (aq) + 2Cl- (aq) PbCl2 (s)• Ag+ (aq)+ Cl- (aq) AgCl (s) • PbCl2 (s) + 2NH3 (aq) + 2H2O (l) Pb(OH-)2 (s) +

2NH4Cl (aq) • AgCl2 (s) + 2NH3 (aq) + 2H2O [Ag(NH3)2]+ (aq) +

2Cl- (aq)+ 2H2O


• When potassium chromate was added to lead nitrate and barium nitrate, Pb(NO3) formed yellow precipitate while barium nitrate formed no precipitate but a yellow-orange solution

reactions involved• Pb2+ + CrO42- PbCrO4• Ba2+ + CrO42- BaCro4• 2CrO4- + 2H+ Cr2O7 + H2O


• After neutralizing, lead nitrate solution formed a white precipitate while barium nitrate solution formed yellow precipitate

reactions involved:

Pb2+ + OH- Pb(OH)2

BaCrO4 Ba2+ + CrO4 2-

Ba2+ CrO4 2- + 2H+ Cr2O7 + H20


• When sodium carbonate was added to barium nitrate it formed a white precipitate. A white precipitate also formed from the reaction of barium sulfate and sodium sulfate but with a faster rate.

Reactions involved:

Ba(NO3)2 + Na2CO3 BaCO3 + 2NaNO3

Ba(NO3)2 + Na2SO4 2NaNO3 + BaSO4


• After adding HCl to the solutions, the barium carbonate precipitate dissolved while the barium sulfate precipitate was still present in the solution.


• The copper nitrate, after adding acetic acid and thioacetamide solution, produced a blue solution and white precipitate.

• After adding nitric acid to one of the test tubes, some of the precipitate were not dissolved

• After adding hydrochloric acid, all of the precipitate was dissolved


Reactions involved:

Cu(NO3)2 + C3H3CSNH2 CuS

CuS Cu2+ + S2-

CuS + 8H+ + 2NO3- 3Cu2+ + 4H2O + 2NO + 3S

Fractional Precipitation

Mix 5 drops of .01 M K2CrO4, Na2S, KI and 20 drops H2O

Add Pb(NO3)2 dropwise

Centrifuge * do after each drop

Record results

Add more Pb(NO3)2 until no more precipitate formation


• Formation of yellow orange precipitate was observed upon the additon lead nitrate to the solution.

• Ideally, 3 colors should be observed with the solution: black (lead sulfate), yellow (lead chromate) and orange (lead iodide).

The reactions involved:

Pb(NO3)2 + S- PbS + 2NO3-

Pb(NO3)2 + CrO4 2- PbCro4 + 2NO3-

Pb(NO3)2 + 2I- PbI2 + 2NO3-

Sources:

• Chemistry 10th Edition by Raymond Chang• http://www.chem.memphis.edu/bridson/FundChem/

T17a1100.htm• http://en.wikipedia.org/wiki/Precipitation_(chemistry)• http://en.wikipedia.org/wiki/Centrifugation

CHEM18.1 Experiment 7_Heterogenous Equilibria

Documents